Question: $\dfrac{ -6f - 3g }{ -3 } = \dfrac{ -4f - 4h }{ -9 }$ Solve for $f$.
Solution: Multiply both sides by the left denominator. $\dfrac{ -6f - 3g }{ -{3} } = \dfrac{ -4f - 4h }{ -9 }$ $-{3} \cdot \dfrac{ -6f - 3g }{ -{3} } = -{3} \cdot \dfrac{ -4f - 4h }{ -9 }$ $-6f - 3g = -{3} \cdot \dfrac { -4f - 4h }{ -9 }$ Multiply both sides by the right denominator. $-6f - 3g = -3 \cdot \dfrac{ -4f - 4h }{ -{9} }$ $-{9} \cdot \left( -6f - 3g \right) = -{9} \cdot -3 \cdot \dfrac{ -4f - 4h }{ -{9} }$ $-{9} \cdot \left( -6f - 3g \right) = -3 \cdot \left( -4f - 4h \right)$ Distribute both sides $-{9} \cdot \left( -6f - 3g \right) = -{3} \cdot \left( -4f - 4h \right)$ ${54}f + {27}g = {12}f + {12}h$ Combine $f$ terms on the left. ${54f} + 27g = {12f} + 12h$ ${42f} + 27g = 12h$ Move the $g$ term to the right. $42f + {27g} = 12h$ $42f = 12h - {27g}$ Isolate $f$ by dividing both sides by its coefficient. ${42}f = 12h - 27g$ $f = \dfrac{ 12h - 27g }{ {42} }$ All of these terms are divisible by $3$ $f = \dfrac{ {4}h - {9}g }{ {14} }$